The electric potential at the surface of an atomic nucleus $(z=50)$ of radius $9 \times 10^{-13} cm$ is $\times 10^{6} V$
Potential $=\frac{k Q}{R}=\frac{k . Z e}{R}$
$=\frac{9 \times 10^{9} \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}}$
$=8 \times 10^{6} V$
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